Started differential geometry part.

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Bokuan Li
2026-01-06 17:25:27 -05:00
parent a704806321
commit a89c447b3b
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\input{./src/topology/index.tex}
\input{./src/fa/index.tex}
\input{./src/measure/index.tex}
\input{./src/dg/index.tex}
\bibliographystyle{abbrv} % We choose the "plain" reference style
\bibliography{refs} % Entries are in the refs.bib file

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\section{Derivatives}
\label{section:derivative}
\begin{definition}[$o(t)$]
\label{definition:little-o}
Let $U \in \cn(0) \subset \real$ and $r: U \to \real$, then $r \in o(t)$ if
\[
\lim_{t \to 0}\frac{o(t)}{t} = 0
\]
\end{definition}
\begin{definition}[Tangent to $0$]
\label{definition:tangent-to-0}
Let $E, F$ be TVSs over $K \in \RC$, $U \in \cn_E(0)$, and $\varphi: U \to F$, then $\varphi$ is \textbf{tangent to $0$} if for any $W \in \cn_F(0)$, there exists $V \in \cn_E(0)$ circled and $r \in o(t)$ such that
\[
\varphi(tV) \subset r(t)W
\]
for sufficiently small $t \in \real$.
\end{definition}
\begin{lemma}
\label{lemma:tangent-to-0}
Let $E, F, G$ be TVSs over $K \in \RC$, $U \in \cn_E(0)$, and $\varphi: U \to F$, $\psi: U \to F$ be tangent to $0$, then:
\begin{enumerate}
\item For any $\lambda \in L(F; G)$, $\lambda \circ \varphi$ is tangent to $0$.
\item $\varphi + \psi$ is tangent to $0$.
\item If $\varphi$ is linear and $F$ is Hausdorff, then $\varphi = 0$.
\end{enumerate}
\end{lemma}
\begin{proof}
(1): Let $W \in \cn_G(0)$, then $\lambda^{-1}(W) \in \cn_F(0)$, and there exists $V \in \cn_E(0)$ and $r \in o(t)$ such that
\[
\lambda \circ \varphi(tV) \subset \lambda (r(t)\lambda^{-1}W) = r(t) \lambda \circ \lambda^{-1} (W) \subset r(t)W
\]
(2): Let $W \in \cn_F(0)$. Using \ref{proposition:tvs-good-neighbourhood-base}, assume without loss of generality that $W$ is circled. Since $\varphi, \psi$ are tangent to $0$, there exists $V_1, V_2 \in \cn_E(0)$ and $r_1, r_2 \in o(t)$ such that
\[
\varphi(tV_1) \subset r_1(t)W \quad \psi(tV_2) \subset r_2(t)W
\]
In which case, since $W$ is circled,
\[
\lambda \varphi(t(V_1 \cap V_2)) + \psi(t(V_1 \cap V_2)) \subset r_1(t) W + r_2(t)W \subset (r_1(t) + r_2(t))W
\]
and $r_1 + r_2 \in o(t)$.
(3): Let $x \in E$ and $W \in \cn_F(0)$. Using \ref{proposition:tvs-good-neighbourhood-base}, assume without loss of generality that $W$ is circled. Since $\varphi$ is tangent to $0$ and linear, then there exists $V \in \cn_E(0)$ and $r \in o(t)$ such that
\[
\varphi(tV) \subset r(t)W \quad \varphi(V) \subset \frac{r(t)}{t}W
\]
for sufficiently small $t \in \real$. Since $W \in \cn_F(0)$, there exists $\lambda \in K$ such that $x \in \lambda V$. Thus
\[
\varphi(x) \in \varphi(\lambda V) = \lambda\varphi(V) \subset \frac{\lambda r(t)}{t}W
\]
As $r \in o(t)$, there exists $t \in \real$ such that $\abs{\lambda r(t)/t} \le 1$, so $\varphi(x) \in \frac{\lambda r(t)}{t}W \subset W$. Since this holds for all $W \in \cn_F(0)$ and $F$ is Hausdorff, $\varphi(x) \in \ol{\bracs{0}} = \bracs{0}$ by \ref{proposition:tvs-closure} and \ref{lemma:t1}.
\end{proof}
\begin{definition}[Derivative]
\label{definition:derivative}
Let $E, F$ be TVSs over $K \in \RC$ with $F$ being Hausdorff, $U \subset E$ open, $f: U \to F$, and $x_0 \in U$, then $f$ is \textbf{differentiable at} $x_0$ if there exists $\lambda \in L(E; F)$, $V \in \cn_E(0)$, and $\varphi: V \to F$ tangent to $0$ such that
\[
f(x_0 + h) = f(x_0) + \lambda h + \varphi(h)
\]
for all $h \in V$. In which case, $Df(x_0) = \lambda$ is the unique continuous linear map satisfying the above, and the \textbf{derivative} of $f$ \textbf{at} $x_0$.
If $f$ is differentiable at every $x_0 \in U$, then $f$ is \textbf{differentiable}, and $Df: U \to L(E; F)$ is the \textbf{derivative} of $f$.
\end{definition}
\begin{proof}
Let $\lambda, \mu \in L(E; F)$ and $\varphi, \psi: V \to F$ be tangent to $0$ such that
\begin{align*}
f(x_0 + h) &= f(x_0) + \lambda h + \varphi(h) \\
&= f(x_0) + \mu h + \varphi(h)
\end{align*}
then
\[
0 = (\lambda - \mu)h + (\varphi - \psi)(h)
\]
By \ref{lemma:tangent-to-0}, $(\lambda - \mu)$ is tangent to $0$ and thus equal to $0$.
\end{proof}
\begin{proposition}[Chain Rule]
\label{proposition:chain-rule}
Let $E, F, G$ be TVSs over $K \in \RC$ with $F, G$ Hausdorff, $U \subset E$ and $V \subset F$ be open, $f: U \to V$ be differentiable at $x_0 \in U$, and $g: V \to G$ be differentiable at $f(x_0)$, then
\[
D(g \circ f)(x_0) = Dg(f(x_0)) \circ f(x_0)
\]
\end{proposition}
\begin{proof}
By differentiability of $f$ and $g$, there exists $\varphi, \psi$ tangent to $0$ such that
\begin{align*}
(g \circ f)(x_0 + h) &= (g \circ f)(x_0) + Dg(f(x_0))[f(x_0 + h) - f(x_0)] + \varphi(f(x_0 + h) - f(x_0)) \\
&= (g \circ f)(x_0) + Dg(f(x_0))[Df(x_0)(h) + \psi(h)] + \varphi(Df(x_0)(h) + \psi(h))
\end{align*}
Since $Dg(f(x_0)) \in L(F; G)$, $Dg(f(x_0)) \circ \psi$ is tangent to $0$ by \ref{lemma:tangent-to-0}.
On the other hand, let $W \in \cn_G(0)$, then there exists $V_1 \in \cn_F(0)$ circled and $r_1 \in o(t)$ such that
\[
\varphi(tV_1) \subset r_1(t)W
\]
for sufficiently small $t \in \real$. Let $V_2 \in \cn_F(0)$ such that $V_2 + V_2 \subset V_1$ and assume without loss of generality that $V_2$ is circled using \ref{proposition:tvs-good-neighbourhood-base}, then there exists $U_1 \in \cn_E(0)$ and $r_2 \in o(t)$ with
\[
\psi(tU_1) \subset r_2(t)V_2
\]
for sufficiently small $t \in \real$. In particular, since $V_2$ is circled, if $t$ is small enough, then
\[
\psi(tU_1) \subset r_2(t)V_2 \subset tV_2
\]
Thus if $U_2 = U_1 \cap Df(x_0)^{-1}(V_2)$, then
\[
\varphi(Df(x_0)(tU_2) + \psi(tU_2)) \subset \varphi(tV_2 + tV_2) \subset \varphi(tV_1) \subset r_1(t)W
\]
so $\varphi(Df(x_0)(h) + \psi(h))$ is tangent to $0$ as well.
\end{proof}

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\chapter{Differential Calculus}
\label{chap:diff}
\input{./src/dg/derivative/derivative.tex}

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src/dg/index.tex Normal file
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\part{Differential Geometry}
\label{part:diffgeo}
\input{./src/dg/derivative/index.tex}